Question

When using the inspection method, the number you would add to (and subtract from) the constant term of the numerator so the polynomial in the numerator will have (x+ 3) as a factor is

Answer 1

Question:

Consider the expression $\frac{x^2 + x -10}{x + 3}$

When using the inspection method the number you would add to (and subtract from) the constant term of the numerator so the polynomial in the numerator will have (x + 3) as a factor is?

Answer:

The constant to add is 4

Step-by-step explanation:

Given

$\frac{x^2 + x -10}{x + 3}$

First, we need to get an expression that has x + 3 has its factor.

Represent this expression with: $(x + 3)(x + k)$

Expand

$x^2 + 3x + kx + 3k$

Group like terms

$x^2 + (3 + k)x + 3k$

Compare the above expression to: $x^2 + x - 10$

$(3 + k)x = x$

$3k = -10$

However, we only consider solving $(3 + k)x = x$ for k

$(3 + k)x = x$

$3 + k = 1$

Subtract 3 from both sides

$3 - 3 + k = 1 - 3$

$k = 1 - 3$

$k= -2$

Substitute -2 for k in $(x + 3)(x + k)$

$(x + 3)(x + k) = (x + 3)(x -2)$

$(x + 3)(x + k) = x^2 + 3x - 2x - 6$

$(x + 3)(x + k) = x^2 + x - 6$

So, the expression that has a factor of x + 3 is $x^2 + x - 6$

To get the constant term to add/subtract, we have:

$Constant = (x^2 + x - 6) - (x^2 + x - 10)$

Open brackets

$Constant = x^2 + x - 6 - x^2 - x + 10$

Collect Like Terms

$Constant = x^2 - x^2+ x - x - 6+ 10$

$Constant = - 6+ 10$

$Constant = 4$